You gotta have faith :-)
\[(x^2 - y^2 + 2yz - z^2) / (x + y - z)\]
Okay, by looking it, we know our answer has to have an x, a y, and a z, right? That's the only way we'll get x^2, y^2 and z^2 terms. We'll do this as a "speculative division" — we'll speculate about the answer, and then see if works. What we do will also bear some similarity to solving multiple equations with combination.
Let's make a little table of what {x, y, z} * (x+y-z) gives us:
\[x(x+y-z) = x^2 + xy - xz\]\[y(x+y-z) = y^2 + xy - yz\]\[z(x+y-z) = -z^2 + xz + yz\]
Those are our building blocks. How do we combine them to get our original numerator?
Well, we need an x in the quotient to produce the x^2 term. We need a -y in the quotient to produce the -y^2 term. We need a z in the quotient to produce the -z^2 term. Let's just add up those equations and see if we get the right thing!
\[x(x+y-z) - y(x+y-z) + z(x+y-z)=\]\[x^2+xy-xz - (y^2+xy-yz) + (-z^2+xz+yz)=\]
\[x^2 + xy - xz -y^2 -xy -(-yz) -z^2 +xz +yz=\]\[x^2-y^2 -z^2 + xy -xy +xz- xz +yz+yz \] and after the dust settles and we cross off all the terms that combine to 0, we have\[x^2-y^2+2yz -z^2\] which is our original numerator. Therefore, the quotient we "guessed", (x - y + z), is correct!